Austria 2014

(a) By the cosine theorem, we have $c^2=a^2+b^2-2ab\cos \gamma$. Since $\cos \gamma >0⟺\gamma <90^{\circ }$, we see that $a^2+b^2>c^2$ is equivalent to $\gamma <90^{\circ }$. Since the analogous results hold for the other two inequalities, we see that these all hold exactly for acute angled triangles $ABC$.

(b) Without loss of generality, assume $c\ge b\ge a$. Since $p=\frac{a}{c}\le 1$ and $q=\frac{b}{c}\le 1$, the three inequalities hold iff $p^n+q^n>1$ holds for all positive integers $n$. If both $p$ and $q$ are less than $1$, there exists sufficiently large values of $n$, such that $p^n<\frac{1}{2}$ and $q^n<\frac{1}{2}$ both hold, and therefore $p^n+q^n<1$. It therefore follows that $q=1$ and $p\le 1$ must hold. The triangle must therefore be isosceles with $b=c$ and $a\le c$, i.e. with $\alpha \le 60^{\circ }$. The inequalities certainly hold for such triangles, and the proof is complete.